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# How to solve this equation for a?

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I'm having difficulty solving the following equation for $$a$$:

$$3a²x-2a³=x³$$

Can anyone help me with the steps? Thanks very much in advance.

Guest Feb 20, 2017

#1
+5

Solve for a:
3 a^2 x - 2 a^3 = x^3

Subtract x^3 from both sides:
-2 a^3 + 3 a^2 x - x^3 = 0

The left hand side factors into a product with three terms:
-(a - x)^2 (2 a + x) = 0

Multiply both sides by -1:
(a - x)^2 (2 a + x) = 0

Split into two equations:
(a - x)^2 = 0 or 2 a + x = 0

Take the square root of both sides:
a - x = 0 or 2 a + x = 0

a = x or 2 a + x = 0

Subtract x from both sides:
a = x or 2 a = -x

Divide both sides by 2:
Answer: |a = x               or                    a = -x/2

Guest Feb 20, 2017
#1
+5

Solve for a:
3 a^2 x - 2 a^3 = x^3

Subtract x^3 from both sides:
-2 a^3 + 3 a^2 x - x^3 = 0

The left hand side factors into a product with three terms:
-(a - x)^2 (2 a + x) = 0

Multiply both sides by -1:
(a - x)^2 (2 a + x) = 0

Split into two equations:
(a - x)^2 = 0 or 2 a + x = 0

Take the square root of both sides:
a - x = 0 or 2 a + x = 0

a = x or 2 a + x = 0

Subtract x from both sides:
a = x or 2 a = -x

Divide both sides by 2:
Answer: |a = x               or                    a = -x/2

Guest Feb 20, 2017
#2
+7002
+5

$$3a^2x-2a^3 = x^3\\ -x^3 + 3a^2x - 2a^3 = 0\\ x^3 - 3a^2x + 2a^3 = 0\\ (x^3-a^3)+(3a^3-3a^2x)=0\\ (x-a)(x^2+ax+a^2)+3a^2(a-x)=0\\ (x-a)(x^2+ax+a^2)-3a^2(x-a)=0\\ (x-a)(x^2+ax-2a^2)=0\\ (x-a)(x^2-a^2+ax-a^2)=0\\ (x-a)((x-a)(x+a)+a(x-a))=0\\ (x-a)(x-a)(x+a+a)=0\\ (x-a)^2(x+2a)=0\\ x-a=0\quad \text{OR}\quad x+2a = 0\\ a = x\quad\text{OR}\quad a=-\dfrac{x}{2}$$

I have more detailed solution on the factorization :)

MaxWong  Feb 20, 2017